Bibliotheca Polyglotta

If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

SI duo triangula duo latera duobus lateribus æqualia habeant, vtrumque vtrique; habeant verò & angulum angulo æqualem sub æqualibus rectis lineis contentum: Et basim basi æqualem habebunt: eritque triangulum triangulo æquale; ac reliqui anguli reliquis angulis æquales erunt, vterque vtrique, sub quibus æqualia latera subtenduntur.

兩三角形。若相當之兩腰線各等。各兩腰線間之角等。則兩底線必等。而兩形亦等。其餘各兩角相當者俱等。

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF. I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

SINT duo triangula A B C, D E F, & vnius vtrumque latus A B, A C, æquale sit alterius vtrique lateri D E, D F, hoc est, A B, ipsi D E, & A C, ipsi D F; angulusqúe A, contentus lateribus A B, A C, æqualis angulo D, contento lateribus D E, D F. Dico basim B C, æqualem quoque esse basi E F; & triangulum A B C, triangulo D E F; & vtrumque angulum B, & C, vtrique angulo E, & F, id est, angulos B, & E, qui opponuntur lateribus æqualibus A C, D F, inter se; & angulos C, & F, qui opponuntur æqualibus lateribus A B, D E, inter se quoque esse æquales.

解曰甲乙丙、丁戊己、兩三角形之甲、與丁、兩角等。甲丙、與丁己、兩線。甲乙、與丁戊、兩線。各等。題言乙丙、與戊己、兩底線必等。而兩三角形亦等。甲乙丙、與丁戊己、兩角。甲丙乙、與丁己戊、兩角。俱等。

For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE. Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF; hence the point C will also coincide with the point F, because AC is again equal to DF. But B also coincided with E; hence the base BC will coincide with the base EF. [For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible. Therefore the base BC will coincide with EF] and will be equal to it. [C.N. 4] Thus the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it. And the remaining angles will also coincide with the remaining angles and will be equal to them, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

Quoniam enim recta A B, rectæ D E, ponitur æqualis, fit, vt si altera alteri superponi intelligatur, collocato puncto A, in puncto D, ¹ ipsæ sibi mutuo congruant, punctumque B, in punctum E, cadat. Neque enim dicere quis poterit, partem rectæ A B, rectæ D E, congruere, & partem non, quia tunc duæ rectæ haberentidem segmentum commune, ² quod est impossibile. Quod si quis dicat, posito puncto A, in D, cadere quidem punctum B, in E, sed rectam A B, cadere vel ad dextram, vel ad sinistram D E, claudent duæ rectæ lineæ superficiem, ³ quod fieri non potest. Quare recta A B, rectæ D E, congruet, vt dictum est. Cum ergo angulus A, angulo D, ponatur æqualis, congruet quoque alter ⁴ alteri, hoc est, recta A C, rectæ D F, congruet, punctumqúe C, in punctum F, cadet, ob æqualitatem rectarum A C, D F. Basis igitur B C, basi E F, congruet quoque: alias si supra caderet, aut infra, vt efficeretur recta E G F, vel E H F, clauderent duæ rectæ E F, E G F; vel E F, E H F, superficiem, (negare enim nemo poterit, tam E G F, quàm E H F, rectam esse, cum vtraque ponatur esse eadem, quæ recta B C.) quod est absurdum. Duæ enim rectæ superficiem ⁵ claudere non possunt. Quocirca ⁶ basis B C, basi E F, æqualis erit, cum neutra alteram excedat; & triangulum A B C, triangulo D E F; & angulus B, angulo E; & angulus C, angulo F, æqualis, ob eandem causam, existet.

1. 8 pronun. 2. 10. pronun. 3. 14. pronun. 4. 8. pronun. 5. 14. pronun. 6. 8. pronun.

論曰。如云乙丙、與戊己、不等。卽令將甲角置丁角之上。兩角必相合、無大小。甲丙、與丁己。甲乙、與丁戊。亦必相合，無大小。公論八此二俱等。而云乙丙、與戊己、不等。必乙丙底或在戊己之上、為庚。或在其下、為辛矣。戊己旣為直線而戊庚己又為直線則兩線當別作一形是兩線能相合為形也辛倣此。公論十二此以非為論者。駁論也。下倣此。

Therefore etc. (Being) what it was required to prove.

Quare si duo triangula duo latera duobus lateribus æqualia habeant, &c. Quod demonstrandum erat.
RECTE Euclides duas conditiones posuit in antecedente huius theorematis, quarum prima est, vt duo latera vnius trianguli æqualia sint duobus lateribus alterius trianguli, vtrumque vtrique; Secunda, vt angulus etiam vnius contentus illis lateribus æqualis sit angulo alterius contento lateribus, quæ is iis sunt æqualia. Deficiente enim alterutra harum conditionum, neque bases, neque reliqui anguli poterunt vnquam esse æquales, vt probe hoc loco à Proclo demonstratur: Triangula verò ipsalicet possint esse æqualia, posteriore duntaxat conditione deficiente, vt ex scholio propos, 37. huius liber constabit, tamen rarò admodum illud continget. Sint enim triangulorum A B C, D E F, anguli A, & D, æquales, nempe recti, & latera A B, A C, æqualia lateribus D E, D F, non quidem vtrumque vtrique, sed illa simul sumpta hisce simul sumptis, sitque A B, 3. A C, 4. vt ambo simul efficiant 7. At verò D E, sit 2. & D E, 5. vt ambo quoque simul 7. constituant. Quibus posisitis, erit basis B C, 5, & b isis E F, radix quadrata huius numeri 29. quæ maior quidem est quam 5. minor autem, quam 6. Item area trianguli A B C, erit 6. area verò trianguli D E F, 5. Anguli denique super basim B C, inæquales erunt angulis super basim E F. Quæ quidem omnia ita esse, hic ostenderemus, nisi adeorum demonstrationem requirerentur multa, quæ nondum sunt confirmata. Vides igitur omnia inæqualia esse, propterea quod non vtrumque latus vtrique lateri æquale existit in dictis triangulis A B C, D E F.
RVRSVS triangulorum A B C, D E F, latera A B, A C, æqualia sint lateribus D E, D F, vtrumque vtrique, sitqúe vnumquodque 5. anguli verò A, & D, contenti dictis lateribus inæquales, sitqúe A, maior quam D. Quibus concessis, erit basis B C, maior base E F, vt propositio 24. huius libri ostendetur. Quod si basim B C. ponamus esse 8. basim autem E F, 4. erit area trianguli A B C, 12. area verò trianguli D E F, radix quadrata huius numeri 84. quæ maior quidem est quam 9. minor verò, quam 10. id quod notissimum est Geometris. Vt igitur duorum triangulorum & bases, & anguli, nec non triangula ipsa æqualia inter se sint, necesse est, vt vtrumque latus vnius æquale sit vtrique lateri alterius, & anguli quoque dictis lateribus contenti æquales existant, vt optimè dixit Euclides.

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